1
$\begingroup$

How do you resolve $\int x (dx)^2$? The notes I am seeing says the term goes to zero, but I don't understand how. Please help.

$\endgroup$
1
  • $\begingroup$ Welcome to Math SE. Could you be a bit more explicit? $\endgroup$
    – mfl
    Nov 15, 2014 at 21:05

2 Answers 2

5
$\begingroup$

Let us interpret $\int_a^b x(dx)^2$ as a limit of Riemann sum. We have $$\int_a^b x(dx)^2 = \lim_{n \to \infty}\sum_{k=1}^{n}\left(a+k\cdot\dfrac{b-a}n\right)\left(\dfrac{b-a}n\right)^2 = \lim_{n \to \infty} \left(an+\dfrac{n(n+1)(b-a)}{2n}\right)\left(\dfrac{b-a}n\right)^2$$ Hence, we obtain $$\int_a^b x(dx)^2 = \lim_{n \to \infty} \left(\dfrac{a(b-a)^2}n+\dfrac{(n+1)(b-a)^3}{2n^2}\right) = 0$$ This is true with any higher order power of $(dx)$, i.e., if $\displaystyle \int_a^b f(x) dx \in \mathbb{R}$, then $\displaystyle \int_a^b f(x) (dx)^p$, where $p>1$ is zero.

$\endgroup$
3
  • $\begingroup$ Very succinct, this helped me understand. Did not even occur to me to represent as Riemann sum. Thank you. $\endgroup$
    – Mark Walsh
    Nov 15, 2014 at 21:15
  • $\begingroup$ I think this argument is invalid. The Riemann integral does not include a general substitution of $dx$ into $\Delta x/n$ for whatever term behind a summation $\int$ symbol. I could claim (fishy as well) that $\int x (dx)^2 = ((1/2) x^2 + C) dx$. $\endgroup$
    – mvw
    Nov 15, 2014 at 21:33
  • $\begingroup$ I think you would be right with that claim, but when you resolve the integral (plug in a and b), the dx term will equal zero, thus the whole equation will equal zero. This is what I think the Riemann sum approach was trying to illustrate. $\endgroup$
    – Mark Walsh
    Nov 17, 2014 at 23:45
1
$\begingroup$

You need to assign a meaning to your intended calculation.

What does $x (dx)^2$ mean in the first place? Is it the same as $x\, dx \,dx$?

What kind of integration is exactly meant if you add an summation $\int$ symbol to your term? What input mathematical objects what output objects what summation procedure?

There is no useful integration for this kind of term defined that I am aware of (maybe some non-standard analysis folk has one).

$\endgroup$
2
  • $\begingroup$ Yes, it does mean x dx dx. The reason I am asking is that I was looking at a proof that used $\int (x^2 + x dx + dx dx)dx = x^3/3$. I found three different sources with the same proof that gave the same answer. I just didn't understand how they got the answer. $\endgroup$
    – Mark Walsh
    Nov 15, 2014 at 21:58
  • $\begingroup$ Looks like 17th century style infinitesimal calculus. So what kind of integral is that supposed to be? Riemann? $\endgroup$
    – mvw
    Nov 15, 2014 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.